Here is the problem by Louis Carrol: make a contur drawing of this
figure by a singular closed line, never passing twice through the
same point. Find all solutions! The first question is: is there
any solution? The answer is - yes, because the contur is an Eulerian
graph: in every vertex there is an even number of edges. Try to
prove this general statement! How many different solutions exist
if the curve may have a finite number of self-intersection
points?

Every Eulerian graph is a projection of some knot or link and
vice versa Such a projection is called regular if the
graph is 4-regular, i.e. if the valence of every vertex is 4.
Otherwise, the projection is irregular. By slightly changing it,
it is always possible to turn some irregular projection of a knot
or link into a regular one. Two knot or link projections are
isomorphic (or simply, equal or same) if they are isomorphic as
the graphs on a sphere. Trying to find all nonisomorphic
projections of alternating knots and links with n
crossings, we need to find all nonisomorphic 4-regular planar
graphs with n vertices and vice versa Among them,
we could distinguish graphs with or without digons. If we denote
digons by colored edges, we could imagine the trefoil as a
triangle with all colored edges, the knot 41 as a
tetrahedron with two colored nonadjacent edges, Borromean rings
as an octahedron...

After that, you could replace every digon
by a chain of digons, and obtain different families of
knots and links. For example, this is the family
generated by the knot 41 If you like that
"geometrical" way of thinking about knots and links,
see the paper
"Geometry of links" by S.V.Jablan.